The Maclaurin series for
f ( x ) = sin ( x ) \displaystyle {f{{\left({x}\right)}}}={\sin{{\left({x}\right)}}} f ( x ) = sin ( x ) is
∑ n = 0 ∞ ( − 1 ) n ( x ) 2 n + 1 ( 2 n + 1 ) ! = x − x 3 3 ! + x 5 5 ! − x 7 7 ! + … \displaystyle {\sum_{{{n}={0}}}^{\infty}}\frac{{{\left(-{1}\right)}^{{n}}{\left({x}\right)}^{{{2}{n}+{1}}}}}{{{\left({2}{n}+{1}\right)}!}}={x}-\frac{{x}^{{3}}}{{{3}!}}+\frac{{x}^{{5}}}{{{5}!}}-\frac{{x}^{{7}}}{{{7}!}}+\ldots n = 0 ∑ ∞ ( 2 n + 1 ) ! ( − 1 ) n ( x ) 2 n + 1 = x − 3 ! x 3 + 5 ! x 5 − 7 ! x 7 + …
By transforming the series for
sin ( x ) \displaystyle {\sin{{\left({x}\right)}}} sin ( x ) , find the Maclaurin series
f ( x ) = x 2 sin ( 7 x ) \displaystyle {f{{\left({x}\right)}}}={x}^{{{2}}}{\sin{{\left({7}{x}\right)}}} f ( x ) = x 2 sin ( 7 x ) .
(Your terms should be entered from smallest degree to highest degree.)
T ( x ) = \displaystyle {T}{\left({x}\right)}= T ( x ) = +
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It would not be easy to write all terms of this series compactly, so we might write this series as:
T ( x ) = ∑ n = 0 ∞ \displaystyle {T}{\left({x}\right)}={\sum_{{{n}={0}}}^{{\infty}}} T ( x ) = n = 0 ∑ ∞ Preview Question 6 Part 6 of 6
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